What is the domain of y=(x^2)/(sqrt(x^2-9)) ?

The domain of y=(x^2)/(sqrt(x^2-9)) is x 3

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domain of y=(x^2)/(sqrt(x^2-9))

domain

y = \frac{x ^{2}}{x ^{2} - 9}

x < - 3 or x > 3

Interval Notation

(- \infty, - 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or x > 3

f (x) = x^{2} - x - ln (x)

f (x) = e^{10 x}

ln (x) + \frac{1}{x ^{2} - 1}

f (x) = 7 + ln (3 x + 1)

f (x) = \frac{x - 5}{x ^{2} - 16}

What is the domain of y=(x^2)/(sqrt(x^2-9)) ?
The domain of y=(x^2)/(sqrt(x^2-9)) is x<-3\lor x>3